$\displaystyle{{x}}^{{3}}-{216}={{x}}^{{3}}-{{6}}^{{3}}\usepackage{cmbright}\Huge\$ , and a binomial in the form $\displaystyle{{a}}^{{3}}-{{b}}^{{3}}\usepackage{cmbright}\Huge\$ can be factored as $\displaystyle{\left({a}-{b}\right)}{\left({{a}}^{{2}}+{a}{b}+{{b}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ . In this problem, $\displaystyle{b}={6}\usepackage{cmbright}\Huge\$ , but that is not the number that is missing. Whenever you factor, remember to check your answer by multiplying the factors back together.